(i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)

Explanation

(i) Power `= +2` dioptre.
(ii) Here, we have `f = +12 cm, R1 = +10 cm, R2 = -15 cm`.
Refractive index of air is taken as unity.
We use the lens formula of Eq. (9.22).
The sign convention has to be applied for `f, R_(1) and R_(2)` .
Substituting the values, we have
`(1)/(12)=(n-1)((1)/(10)-(1)/(-15))`
This gives `n=1.5`.
(iii) For a glass lens in air, `n_(2)=1.5, n_(1)=1, f=+20cm`. Hence, the lens formula gives
`(1)/(20)=0.5[(1)/(R_(1))-(1)/(R_(2))]`
For the same glass lens in water, `n_(2)=1.5, n_(1)=1.33`. Therefore, `(1.33)/(f)=(1.5-1.33)[(1)/(R_(1))-(1)/(R_(2))]`
Combining these two equations, we find `f=+78.2cm`.